Whakaoti i te 3s^2+16s+5

$$3 s ^ { 2 } + 16 s + 5$$

Tauwehe

$$\left(s+5\right)\left(3s+1\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 3s^{2}+as+bs+5. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=16 ab=3\times 5=15$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōrunga te a+b, he tōrunga hoki a a me b. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 15.

$$1,15 3,5$$

Tātaihia te tapeke mō ia takirua.

$$1+15=16 3+5=8$$

Ko te otinga te takirua ka hoatu i te tapeke 16.

$$a=1 b=15$$

Tuhia anō te 3s^{2}+16s+5 hei \left(3s^{2}+s\right)+\left(15s+5\right).

$$\left(3s^{2}+s\right)+\left(15s+5\right)$$

Tauwehea te s i te tuatahi me te 5 i te rōpū tuarua.

$$s\left(3s+1\right)+5\left(3s+1\right)$$

Whakatauwehea atu te kīanga pātahi 3s+1 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(3s+1\right)\left(s+5\right)$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Ka taea te huamaha pūrua te tauwehe mā te whakamahi i te huringa ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), ina ko x_{1} me x_{2} ngā otinga o te whārite pūrua ax^{2}+bx+c=0.

$$3s^{2}+16s+5=0$$

Ko ngā whārite katoa o te āhua ax^{2}+bx+c=0 ka taea te whakaoti mā te whakamahi i te tikanga tātai pūrua: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. E rua ngā otinga ka puta i te tikanga tātai pūrua, ko tētahi ina he tāpiri a ±, ā, ko tētahi ina he tango.

$$s=\frac{-16±\sqrt{16^{2}-4\times 3\times 5}}{2\times 3}$$

Pūrua 16.

$$s=\frac{-16±\sqrt{256-4\times 3\times 5}}{2\times 3}$$

Whakareatia -4 ki te 3.

$$s=\frac{-16±\sqrt{256-12\times 5}}{2\times 3}$$

Whakareatia -12 ki te 5.

$$s=\frac{-16±\sqrt{256-60}}{2\times 3}$$

Tāpiri 256 ki te -60.

$$s=\frac{-16±\sqrt{196}}{2\times 3}$$

Tuhia te pūtakerua o te 196.

$$s=\frac{-16±14}{2\times 3}$$

Whakareatia 2 ki te 3.

$$s=\frac{-16±14}{6}$$

Nā, me whakaoti te whārite s=\frac{-16±14}{6} ina he tāpiri te ±. Tāpiri -16 ki te 14.

$$s=-\frac{2}{6}$$

Whakahekea te hautanga \frac{-2}{6} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$s=-\frac{1}{3}$$

Nā, me whakaoti te whārite s=\frac{-16±14}{6} ina he tango te ±. Tango 14 mai i -16.

$$s=-\frac{30}{6}$$

Whakawehe -30 ki te 6.

$$s=-5$$

Tauwehea te kīanga taketake mā te whakamahi i te ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Me whakakapi te -\frac{1}{3} mō te x_{1} me te -5 mō te x_{2}.

$$3s^{2}+16s+5=3\left(s-\left(-\frac{1}{3}\right)\right)\left(s-\left(-5\right)\right)$$

Whakamāmātia ngā kīanga katoa o te āhua p-\left(-q\right) ki te p+q.

$$3s^{2}+16s+5=3\left(s+\frac{1}{3}\right)\left(s+5\right)$$

Tāpiri \frac{1}{3} ki te s mā te kimi i te tauraro pātahi me te tāpiri i ngā taurunga. Ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$3s^{2}+16s+5=3\times \frac{3s+1}{3}\left(s+5\right)$$

Whakakorea atu te tauwehe pūnoa nui rawa 3 i roto i te 3 me te 3.

$$3s^{2}+16s+5=\left(3s+1\right)\left(s+5\right)$$

Aromātai

$$\left(s+5\right)\left(3s+1\right)$$